Chapter 4.6, Problem 20PPS

To determine

To find: the inverse of the matrix, if it exists.

Answer to Problem 20PPS

The inverse exists.

Explanation of Solution

Given information:

  $\left[\begin{array}{cc}1& -1\\ -6& -1\end{array}\right]$ .

Formula used:

Inverse of a $2\times 2$ matrix $A=\left[\begin{array}{cc}a& b\\ c& d\end{array}\right]$ is $A^{-1}=\frac{1}{ad-bc}\left[\begin{array}{cc}d& -b\\ -c& a\end{array}\right]$ is, where $ad-bc\ne 0$ .

The value of a second order determinant is the difference of the products of the products of the two diagonals.

Calculation:

Consider,

  $P=\left[\begin{array}{cc}1& -1\\ -6& -1\end{array}\right]$

Step 1.

  $=\left|\begin{array}{cc}1& -1\\ -6& -1\end{array}\right|=\left(1\times -1\right)-\left(-1\times -6\right)$

  $\begin{array}{l}=-1-6\\ =-7\end{array}$

Since the determinant does not equal to 0, $P^{-1}$ exists.

Step 2.

  $P^{-1}=\frac{1}{(1)(-1)-(-1)(-6)}\left[\begin{array}{cc}-1& 1\\ 6& 1\end{array}\right]$

  $\begin{array}{l}=\frac{1}{-1-6}\left[\begin{array}{cc}-1& 1\\ 6& 1\end{array}\right]\\ =-\frac{1}{7}\left[\begin{array}{cc}-1& 1\\ 6& 1\end{array}\right]\end{array}$

  $a=-1,b=1,c=6,d=1$

Step3.

Thus, inverse exits.

  $P^{-1}=\left[\begin{array}{cc}\frac{1}{7}& -\frac{1}{7}\\ -\frac{6}{7}& -\frac{1}{7}\end{array}\right]$